Assignment 7 Query

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course Mth 151

11/19 1

007. `Query 7

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Question: `qQuery 1.2.18 (previously 1.2.6) seq 2, 57, 220, 575, 1230, 2317 ... by successive differences

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Your solution:

I am not sure how to do this or how to begin doing this.

confidence rating #$&*:

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Given Solution:

`a** Using sequences of differences we obtain:

2, 57, 220, 575, 1230, 2317, # 3992

55, 163, 355, 655, 1087, # 1675

108, 192, 300, 432, # 588

84, 108, 132, # 156

24, 24,

The final results, after the # signs, are obtained by adding the number in the row just below, in the following order:

Line (4) becomes 132+24=156

Line (3) becomes 432+156=588

Line (2) becomes 1087+588=1675

Line (1) becomes 2317+1675=3992

The next term is 3992. **

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Self-critique (if necessary):

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&#You did not answer the given question. You need to always at least explain what you do and do not understand about the question. A phrase-by-phrase analysis is generally required when you cannot otherwise answer a question.

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Self-critique Rating:1

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Question: `q 1.2.30 (previously 1.2.18) 1^2 + 1 = 2^2 - 2; 2^2 + 2 = 3^2 - 3; 3^2 + 3 = etc.

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Your solution:

3^2+3=4^4-4;

confidence rating #$&*:1

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Given Solution:

`a** The next equation in the sequence would be 4^2 + 4 = 5^2 - 5

The verification is as follows:

4^2 + 4 = 5^2 - 5 simplifies to give you

16 + 4 = 25 - 5 or

20 = 20 **

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Self-critique (if necessary):

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Self-critique Rating:1

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Question: `q1.2.42 (previously 1.2.30) state in words (1 + 2 + ... + n ) ^ 2 = 1^3 + 2^3 + ... + n^3

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Your solution:

If it was (1+2)^2 = 1^3+2^3

(1+2)^2=9

And

1^3+2^3=9

Since this is correct then above would be as well

confidence rating #$&*:3

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Given Solution:

`a** the equation says that the square of the sum of the first n counting numbers is equal to the sum of their cubes **

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Self-critique (if necessary):

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Self-critique Rating:3

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Question: `q (previously 1.2.36)

Divide the first triangular number by 3 and write down the remainder.

Divide the second triangular number by 3 and write down the remainder.

Divide the third triangular number by 3 and write down the remainder.

Divide the fourth triangular number by 3 and write down the remainder.

Continue until you have established a pattern.

What is the pattern?

Can you explain why the pattern occurs? Is it possible that the pattern doesn't go on forever?

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Your solution:

.33, 1, 2, 3.33,5 ,7, 9.33333 every two numbers the next ends with .3333repeating

Not sure why, I can see it being infinite

confidence rating #$&*:2

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Given Solution:

`a** The triangular numbers are 1, 3, 6, 10, 15, 21, . . . . We divide these by 3 and get the sequence of remainders.

When you divide 1 by 3 you get 0 with remainder 1 (3 goes into 1 zero times with 1 left over). 3 divided by 3 gives you 1 with remainder 0. 6 divded by 3 is 2 with remainder 0. 10 divided by 3 is 3 with remainder 1. Therefore the remainders are 1,0,0,1,0,0.

It turns out that the sequence continues as a string of 1,0,0 's. At this point that is an inductive pattern, but remmeber that the sequence of triangular numbers continues by adding successively larger and larger numbers to the members of the sequence. Since the number added always increases by 1, and since every third number added is a multiple of 3, is isn't too difficult to see how the sequence of remainders comes about and to see why it continues as it does.

COMMON ERROR: .3333333,1,2,3.3333333,etc.

INSTRUCTOR CORRECTION:

You need the remainders, not the decimal equivalents. When you divide 1 by 3 you get 0 with remainder 1 (3 goes into 1 zero times with 1 left over). 3 divided by 3 gives you 1 with remainder 0. 6 divded by 3 is 2 with remainder 0. 10 divided by 3 is 3 with remainder 1. Therefore the remainders are 1,0,0,1,0,0 and the sequence continues as a string of 1,0,0 's.

COMMON ERROR: 1/3, 1, 2, 3 1/3

CORRECTION:

These are the quotients. You need the remainders. If you get 1/3 that means the remainder is 1; same if you get 3 1/3. If you just getting whole number (like 1 or 2 in your calculations) the remainder is 0.

In other words, when you divide 1 by 3 you get 0 with remainder 1 (3 goes into 1 zero times with 1 left over). 3 divided by 3 gives you 1 with remainder 0. 6 divded by 3 is 2 with remainder 0. 10 divided by 3 is 3 with remainder 1.

The remainders form a sequence 1,0,0,1,0,0 and the sequence continues as a string of 1,0,0 's. **

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Self-critique (if necessary):

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&#You need a detailed self-critique here.

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Self-critique Rating:2

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Question: `q 1.2. 58 (previously 1.2.48) use formula to find the 12 th octagonal number. Explain in detail how you used the formula to find this number.

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Your solution:

12(2)(6(12-4))

24(6(8))

24(48)

1152

confidence rating #$&*:3

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Given Solution:

`a** The pattern to formulas for triangular, square, pentagonal, hexagonal, heptagonal and octagonal numbers is as follows (you should know this pattern; you might need one or more of these formulas on the test):

Triangular numbers: n / 2 * [ n + 1 ] note that this is the same as Gauss' formula

Square numbers: n / 2 * [ 2n + 0 ] or just n^2

Pentagonal #'s: n / 2 * [ 3n - 1 ]

Hexagonal #'s: n / 2 * [ 4n - 2 ]

Heptagonal #'s: n / 2 * [ 5n - 3 ]

Octagonal #'s: n / 2 * [ 6n - 4 ]

The coefficient of n in the bracketed term starts with 1 and increases by 1 each time, and the +1 in the first bracketed term decreases by 1 each time.

You will need to know these formulas for the test.

The last formula is for octagonal numbers. To get n = 12 octangonal number use n/2 * [ 6n - 4 ] to get 12 / 2 * [ 6 * 12 - 4 ] = 6 * 68 = 408. **

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&#This also requires a self-critique.

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Self-critique (if necessary):

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Self-critique rating:

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Question: `q 1.2. 58 (previously 1.2.48) use formula to find the 12 th octagonal number. Explain in detail how you used the formula to find this number.

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Your solution:

12(2)(6(12-4))

24(6(8))

24(48)

1152

confidence rating #$&*:3

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Given Solution:

`a** The pattern to formulas for triangular, square, pentagonal, hexagonal, heptagonal and octagonal numbers is as follows (you should know this pattern; you might need one or more of these formulas on the test):

Triangular numbers: n / 2 * [ n + 1 ] note that this is the same as Gauss' formula

Square numbers: n / 2 * [ 2n + 0 ] or just n^2

Pentagonal #'s: n / 2 * [ 3n - 1 ]

Hexagonal #'s: n / 2 * [ 4n - 2 ]

Heptagonal #'s: n / 2 * [ 5n - 3 ]

Octagonal #'s: n / 2 * [ 6n - 4 ]

The coefficient of n in the bracketed term starts with 1 and increases by 1 each time, and the +1 in the first bracketed term decreases by 1 each time.

You will need to know these formulas for the test.

The last formula is for octagonal numbers. To get n = 12 octangonal number use n/2 * [ 6n - 4 ] to get 12 / 2 * [ 6 * 12 - 4 ] = 6 * 68 = 408. **

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&#This also requires a self-critique.

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Self-critique (if necessary):

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Self-critique rating:

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&#Your work looks good. See my notes. Let me know if you have any questions. &#